1. Let R = C[x].
(a) Describe all saturated multiplicative subsets S in R.
(b) For each such S, describe the localization S −1 R.
(c) For which S is S −1 R a finitely generated C-algebra? A local
algebra?
(d) What are the S-saturated ideals of R (for each S)? What is the
saturation of the non-saturated ones?
(e) For any finitely generated module M over R and any S, describe
−1
S M.
Solution. (a) Let S be a saturated multiplicative subset. If 0 ∈ S
then any x ∈ S, i.e. S = R, since 0x = 0. Assume 0 ∈
/ S. Every
invertible element belongs to S, so C ⊂ S. If
p = a0 (x − x1 )n1 ...(x − xk )nk ∈ S
for ni > 0 then x − xi ∈ S. So S = S(Z) is determined by the set Z of
all a ∈ C such that x − a ∈ S; namely, S consists of polynomials p as
above such that roots of p are in Z. Clearly, any Z can arise.
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(b) 0−1 R = 0, S(Z)−1 R = C[x, x−a
, a ∈ Z] is the ring of rational
finctions which are regular outside of Z.
(c) S(Z)−1 R is finitely generated as an algebra iff Z is finite. Otherwise, we can create only finitely many poles. To have a local ring, we
need Z to be everything or miss just one point.
(d) Each ideal I is principal, generated by p ∈ C[x]. Consider the
nontrivial case p 6= 0. Saturated means that sp ∈ I for some s ∈ S =
S(Z) means p ∈ I. This is equivalent to saying that the roots of p are
disjoint from Z. The saturation of I in general is I 0 , generated by the
polynomial q such that p = qf with f having roots in Z and q having
no roots in Z.
(e) A finitely generated module is Rm ⊕ E, where E is finite dimensional. So it suffices to describe S −1 E. Recall that E is a direct sum
of Jordan block modules. Multiplication by S −1 just knocks out the
Jordan blocks that have eigenvalue in Z.
2. Let R = C[x, y]/(y 2 − x(x − 1)(x − 2)).
(a) Describe X = SpecR (and draw a picture of the real points).
(b) Describe the open sets in X in Zariski topology.
(c) Let P be any point on the curve y 2 = x(x − 1)(x − 2), and M be
the ideal consisting of f ∈ R such that f (P ) = 0. Show that M is a
projective (=locally free) R-module of rank 1, but it is not free.
(d) (harder!) Which of the open sets on X are principal? For which n
is there a set of distinct points P1 , ..., Pn on the curve y 2 = x(x−1)(x−2)
such that X \ {P1 , ..., Pn } is not principal?
Hint. You may need to use the addition law on the elliptic curve.
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Solution.
(a) X = SpecR is the union of the curve C defined by the equation
2
y = x(x − 1)(x − 2) (in C2 ) and the “generic” point g corresponding
to the ideal 0.
(b) Nonempty open sets are complements of finite sets in C.
(c) An ideal is a free module iff it is a principal ideal, so we must
show that M is not principal, i.e. there is no regular function f ∈ R
such that M = (f ). For that f would have to have a simple zero at P
and no other zeros on C. Let us show that there is no such function.
Since y 2 expresses in terms of x, any f has the form a(x)y +b(x), where
a, b are polynomials. If a = 0 and f = b(x) is not a constant then f
has at least two zeros or one double zero. If a 6= 0, then the solutions
of f = 0 on C are found from the system
a(x)y + b(x) = 0, y 2 = x(x − 1)(x − 2).
Thus we get
a(x)2 x(x − 1)(x − 2) = b(x)2 .
This equation has degree at least 3 (otherwise the leading term on
the left side has to be canceled by something, which can’t happen, as
RHS is of even degree). So we will have at least three roots, counting
multiplicities. This means that we cannot arrange for just one simple
zero. So M is not free.
On the other hand, M is locally free of rank 1. To show this, for any
maximal ideal z ∈ C, z 6= P , pick a regular function f ∈ R so that
f (z) 6= 0 and f (P ) = 0. Then Mf = Rf . Also, MP is the maximal
ideal in RP , so it is free over RP (as it is a principal ideal generated by
any f ∈ R such that f has a simple zero at P ).
(d) Recall that we have an abelian group law on C ∪ {∞} (with ∞
playing the role of 0 in the group), defined as follows. If P, Q are points
on C then to compute P + Q, draw a line through P, Q and take R to
be the third intersection point of this line with C. Then P + Q = −R
is the point mirror symmetric to R with respect to the x-axis. (We
interpret intersections with multiplicities; e.g., if P = Q, we draw the
tangent).
The complement U of points P1 , ..., Pn is a principal open set iff there
exists a regular function f on C such that f vanishes at P1 , ..., Pn and
nowhere else. Let m1 , ..., mn be the multiplicitiesP
of f at P1 , ..., Pn . We
claim that f with such multiplicities exists iff i mi Pi = 0 (for the
above addition law).
Let us first show that this condition is sufficient. To this end, let
P, Q, R be three points such that P + Q + R = 0, and let gP,Q,R be
the linear function defining the line through P, Q, R. Then gP,Q,R has
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simple zeros at P, Q, R and no other zeros. Also,P
x − xQ has simple
zeros at Q and −Q. Now we can do induction in
mi . If n > 1, let
P1 + P2 = Q. Also let f be the function vanishing m1 − 1 times at P1 ,
m2 − 1 times at P2 , mi times at Pi for i ≥ 3, and another time at Q.
By the induction assumption f exists. Then the function
gP1 ,P2 ,−Q f
x − xQ
is regular and has the required pattern of zeros. If n = 1, we set
2P1 = Q, and do the same thing.
P
To prove that the condition is necessary, suppose mi Pi = −P 6= 0,
and let f be a function with such pattern of zeros. Then by sufficiency
proved above, there is g which has the required pattern of zeros and
another zero at P . Then g/f is a regular function with a single zero at
P , which does not exist by (c), a contradiction.
PSo we see that for U to be principal, there must exist mi such that
mi Pi = 0. This does not hold in general; e.g. if n = 1 and P1 = P ,
this means that P is a finite order element in the group, and it is easy
to see that there are countably many elements of finite order, while C
is uncountable. In fact, we can see that for any n there are P1 , ..., Pn
such that the complement of P1 , ..., Pn is not principal.
(Here I am assuming to be known that a rational function on C is
regular iff it has no poles at C; this is not hard to show.)
3. Let R = C[x1 , ..., xr ], and φ : Rm → Rn is a module map, given by
a matrix [φ] = (aij ), aij ∈ R. Let M = Cokerφ. Describe the support
of M explicitly (as V (a), where the ideal a is defined by polynomial
equations). Consider first m < n, then m = n, then m > n.
Solution. If m < n, the support is the entire SpecR, so let us
consider the case m ≥ n. A linear map Cm → Cn (m ≥ n) given by a
matrix A = (aij ) is not surjective iff all n by n minors of A are zero.
So let us show that the ideal a is generated by the n by n minors of A
(as polynomials of x1 , ..., xr ). To this end, assume first that a prime p
contains a, i.e., the minors are zero in R0 := R/p. Then the minors are
also zero over Frac(R0 ), so the map is not surjective after localization
to the field of fractions. This implies that the map
(1)
(Rp /pRp )m → (Rp /pRp )n
is not surjective, and hence
(2)
Rpm → Rpn
is not surjective, i.e. Mp 6= 0, so p ∈ SuppM . On the other hand,
assume that p does not contain some minor ∆ of A. Then ∆ 6= 0 in R0 ,
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so the map is surjective over Frac(R0 ). Hence the map (1) is surjective.
By Nakayama lemma, this implies that the map (2) is surjective, hence
Mp = 0.
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